Understanding Serre-Chevalley relations
The relations prescribe how the Lie algebra is supposed to decompose when considered as a module over the copy ${\mathfrak s}{\mathfrak l}_2(i)$ of ${\mathfrak s}{\mathfrak l}_2({\mathbb k})$ spanned by $\{e_i,f_i,h_i\}$. Namely, if you know that $\text{ad}(e_i)^{a+1}(e_j)=0$ but $\text{ad}(e_i)^{a}(e_j)\neq 0$, then the ${\mathfrak s}{\mathfrak l}_2(i)$-submodule of ${\mathfrak g}$ spanned by $e_j$ has dimension $a+1$ (note that $\text{ad}(f_i)(e_j)=0$, so $e_j$ is a lowest weight vector for the generated ${\mathfrak s}{\mathfrak l}_2(i)$ submodule).
If you look at the A2 root system of ${\mathfrak s}{\mathfrak l}_3({\mathbb C})$ for example, you see that if $\{\alpha,\beta\}$ is a basis of the root system, then the root string $\alpha, \alpha + \beta, ...$ has only length $2$, in accordance with the fact that the Cartan matrix is $\tiny\begin{pmatrix} 2 & -1 \\ -1 & 2\end{pmatrix}$. If, in contrast, you look at the G2 root system, you'll see one chain of length $4$ and one of length $2$, in accordance with the Cartan matrix $\tiny\begin{pmatrix} 2 & -3 \\ -1 & 2\end{pmatrix}$. The last Serre-Chevallley relation reflects these chain lengths (even the $2$'s on the diagonal make sense, because the ${\mathfrak s}{\mathfrak l}_2(i)$ submodule spanned by $e_i$ is just ${\mathfrak s}{\mathfrak l}_2(i)$ itself, so has dimension $3$; the sign is different because $e_i$ is a highest weight vector, though).
It might help if we label the generators by the corresponding roots instead (since this is where they come from). When we do this, we get that if $[e_{\alpha_i},e_{\alpha_j}]$ is in the root space corresponding to $\alpha_i + \alpha_j$, so since there are only finitely many roots, at some point, if we keep taking brackets with the same $e_{\alpha_i}$, we need to get $0$.
That was the reason coming from actually looking at the Lie algebra we know we should get. But there is another way to look at this: What would happen if we left these out?
As it happens, if we leave out the relation, we no longer get a finite dimensional Lie algebra, so what the relation really does is make sure the Lie algebra stays small.